Question

The value of the integral $\int_\limits{-\pi/4}^{\pi/4} \log (\sec \theta-\tan \theta) d \theta$ is

• A. $\frac {\pi}{4}$
• B. $\frac {\pi}{2}$
• C. $0$
• D. $\pi$

Answer 1

The Correct Option is C

Let $I=\int_\limits{-\pi / 4}^{\pi / 4} \log (\sec \theta-\tan \theta) d \theta$
Again, let $f(\theta)=\log (\sec \theta-\tan \theta)$
$\therefore f(-\theta)=\log [\sec (-\theta)-\tan (-\theta)]$
$=\log \left[(\sec \theta+\tan \theta) \times \frac{\sec \theta-\tan \theta}{\sec \theta-\tan \theta}\right]$
$=\log \left[\frac{\sec ^{2} \theta-\tan ^{2} \theta}{\sec \theta-\tan \theta}\right]=\log \left[\frac{1}{\sec \theta-\tan \theta}\right]$
$=\log 1-\log (\sec \theta-\tan \theta)$
$=0-\log (\sec \theta-\tan \theta)$
$\Rightarrow f(-\theta)=-f(\theta)$
Hence, $f(\theta)$ is an odd function.
$\therefore I=0$